Question

Find the sum to n terms
i) 0.9+0.99+0.999+...
ii) 0.5+0.55+0.555+...

Answer 1

1, 0.9+0.99+0.999 = 2.889
2, 0.5+0. 55+0.555 = 1.805

Answer 2

Answer:

i)$n - [ 1 -(0.1)^n]$

ii)$n - [ 1 -(0.1)^n]$

Step-by-step explanation:

We have to find the sum of the GP

i) 0.9+0.99+0.999+...

(1 - 0.1)+(1 - 0.01)+(1 - 0.001)+(1 - 0.0001)+..........

= (1+1+1+1+....... n) +( -0.1-0.01-0.001- .........)

= (1 × n) - (0.1 + 0.01 + 0.001 + 0.0001 + .........) ....(1)

Formula of geometric sum upto n terms, where a = first term, r  = common ratio of GP:

$\dfrac{a(1 - r^n)}{1- r}$

0.1 + 0.01 + 0.001 + 0.0001 + .........

a = 0.1, r = 0.1

$S_n = \dfrac{0.1(1 - (0.1)^n)}{1- 0.1} =\dfrac{0.1(1 - (0.1)^n)}{0.9}\\S_n = \dfrac{(1 - (0.1)^n)}{9}$

Put the value in equation (1)

$= n - \dfrac{(1 - (0.1)^n)}{9}$ is the sum of the given GP

ii) 0.5+0.55+0.555+...

5(0.1 + 0.01 + 0.001 + ...........) ....(1)

Now multiply equation (1) by 9/9

5/9(0.9+0.99+0.999+........)

5/9[(1 - 0.1)+(1 - 0.01)+(1 - 0.001)+(1 - 0.0001)+..........]

5/9[(1+1+1+1+....... n) +( -0.1-0.01-0.001- .........)]

5/9[(1 × n) - (0.1 + 0.01 + 0.001 + 0.0001 + .........)] ....(2)

Formula of geometric sum upto n terms, where a = first term, r  = common ratio of GP:

$\dfrac{a(1 - r^n)}{1- r}$

0.1 + 0.01 + 0.001 + 0.0001 + .........

a = 0.1, r = 0.1

$S_n = \dfrac{0.1(1 - (0.1)^n)}{1- 0.1} =\dfrac{0.1(1 - (0.1)^n)}{0.9}\\S_n = \dfrac{(1 - (0.1)^n)}{9}$

Put the value in equation (2)

$\dfrac 59[ n - \dfrac{(1 - (0.1)^n)}{9}]$ is the sum of the given GP